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Thread: Coordinate Geometry

  1. #1
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    Coordinate Geometry



    In the diagram above, triangle $\displaystyle ACD$ is such that the coordinates of $\displaystyle A$ and $\displaystyle B$ are $\displaystyle (2, 12)$ and $\displaystyle (6, 9)$ respectively. $\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC $. The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$. The equation of $\displaystyle AD$ is $\displaystyle 3y=5x+26$.

    Calculate the coordinates of C.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let $\displaystyle M_1(x_1,y_1), \ M_2(x_2,y_2)$ be two points and $\displaystyle M(x_M,y_M)$ a point on the segment $\displaystyle M_1M_2$ such as $\displaystyle \displaystyle\frac{M_1M}{MM_2}=k$.

    Then $\displaystyle x_M=\displaystyle\frac{x_1+kx_2}{1+k}, \ y_M=\displaystyle\frac{y_1+ky_2}{1+k}$

    In this case $\displaystyle \displaystyle\frac{AB}{AC}=\frac{1}{2}$

    Then $\displaystyle x_B=\displaystyle\frac{x_A+\frac{1}{2}x_C}{1+\frac {1}{2}}$

    $\displaystyle y_B=\displaystyle\frac{y_A+\frac{1}{2}y_C}{1+\frac {1}{2}}$

    You know the coordinates of A and B, so you can find the coordinates of C.
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  3. #3
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    Hello, Punch!

    I assume there are more parts to this problem.
    There is way too much information . . .


    In the diagram, $\displaystyle \Delta ACD$ has points: $\displaystyle A(2,12)$ and $\displaystyle B(6,9).$

    $\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC $.
    The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$.
    The equation of $\displaystyle AD$ is: $\displaystyle 3y\:=\:5x+26$.

    Calculate the coordinates of $\displaystyle C.$

    Here is a simplified diagram:


    Code:
          |
          |       A
          | (2,12)o - - +
          |        \    :
          |         \   :
          |          \  :
          |           \ :
          |            \:
          |       B(6,9)o - - - - +
          |              \        :
          |               \       :
          |                \      :
          |                 \     :
          |                  \    :
          |                   \   :
          |                    \  :
          |                     \ :
          |                      \:
          |                     C o
          |
      - - + - - - - - - - - - - - - - -
          |

    $\displaystyle \text{Since }BC = \frac{2}{3}AC,\; BC\text{ is twice }AB$

    Going from $\displaystyle A$ to $\displaystyle B$, we move: 4 units right and 3 units down.

    So, going from $\displaystyle B$ to $\displaystyle C$, we move: 8 units right and 6 units down.


    Therefore, point $\displaystyle C$ is $\displaystyle (14,3).$

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  4. #4
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    you are right soroban
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